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3.6 Principal Amortization and Interest Payment of a Loan
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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption
Repayment Method 3: the principal sum borrowed to be paid in a lump sum at the end of Year 5 is as shown in Table 3.3.
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Table 3.3 – Single principal amortization for a loan of $100,000
The present value of the annual installments for this method will of course be the same again, i.e. $100,000, if i is taken as 5 percent per annum. The verification is left to the readers.
We should pay attention to columns (3) and (4) of Table 3.3. Apparently, the annual interest seems like simple interest calculation, giving $5,000 of interest constantly from Year 1 to Year 5. The calculation has been based on the multiplication of the principal sum by the interest rate (i.e. P × i = $100,000 × 5% per annum), which looks similar to the method of simple interest calculation.
However, the interest payment shown in Table 3.3 is in fact calculated by compound interest principle, with the new balances (i.e. column 3) remain unchanged at $100,000, the same as the principal, over the whole period of loan recovery of five years. In such a case, the interest is actually charged annually to the new balance (i.e. compound interest principle), as if it is charged to the principal sum.
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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption a) How much should the company pay at the end of each year to the bank (assuming uniform
payment)?
b) The bank changes the interest rate to 4% p.a. at the beginning of the third year. What will be the amount of the company’s last payment (i.e. payment at the end of year 5) if it keeps on paying the bank the same amount as calculated in (a) above at the end of years 3 and 4?
c) Continued from (b), what will be the company’s repayment schedule if it chooses to pay back the bank in the form of uniform payments at the end of years 3, 4 and 5?
Answer:
a) Yearly payment = 1,000,000 ×
() ()
− +
+ 1 1
1
n n
i i
i (for i = 0.05 and n = 5)
= 100,000 × 0.2309748
= $23,097.48
End of year
Principal unpaid Yearly payment Interest Principal amortization
Remaining principal
1 $100,000.00 $23,097.48 $5,000.00 $18,097.48 $81,902.52
2 $81,902.52 $23,097.48 $4,095.12 $19,002.36 $62,900.16
3 $62,900.16 $23,097.48 $3,145.01 $19,952.47 $42,947.69
4 $42,947.69 $23,097.48 $2,147.38 $20,950.10 $21,997.59
5 $21,997.59 $23,097.48 $1,099.89 $21,997.59 $0.00
In the above table, for year 1, interest = 100,000 × 5% = 5,000. The principal amortization = 23,097.48 – 5,000
= 18,097.48, and therefore the remaining principal = 100,000 – 18,097.48 = 81,902.52. For year 2, interest = 81,902.52 × 5% = 4,095.12, and the principal amortization = 23,097.48 – 4,095.12 = 19,002.36, and hence the remaining principal = 81,902.52 – 19,002.36 = 62,900.16. Similar calculations are applied to years 3, 4 and 5. The final remaining principal would naturally be equal to zero at the end of year 5.
It is interesting to note that the five figures under the column Principal amortization sum up to exactly the loan amount $100,000. The same is true for parts (b) and (c) below.
b) In the part (b) table, only the first two rows (years 1 and 2) are similar to the part (a) table (5% p.a. interest rate). For years 3, 4 and 5, interest = principal unpaid × 4%.
End of year Principal unpaid Yearly payment Interest Principal amortization
Remaining principal
1 $100,000.00 $23,097.48 $5,000.00 $18,097.48 $81,902.52
2 $81,902.52 $23,097.48 $4,095.12 $19,002.36 $62,900.16
3 $62,900.16 $23,097.48 $2,516.01 $20,581.47 $42,318.69
4 $42,318.69 $23,097.48 $1,692.75 $21,404.73 $20,913.96
5 $20,913.96 $21,750.52 $836.56 $20,913.96 $0.00
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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption So, if the company keeps on paying the bank the same amount as calculated in (a) at the end of years 3
& 4, the amount of the company’s last payment is $21,750.52 ($836.56 interest already included).
c) Because the interest rate changes to 4% p.a. at the beginning of the 3rd yr, the new uniform payment is:
$62,900.16 ×
()
()
− +
+ 1 1
1
n n
i i
i (for i = 0.04 and n = 3)
= $62,900.16 × 0.3603485
= $22,665.98
If the company chooses to pay back the bank in the form of uniform payments at the end of years 3, 4
& 5, the repayment schedule will be like this:
End of year Principal unpaid Yearly payment Interest Principal amortization
Remaining principal
1 $100,000.00 $23,097.48 $5,000.00 $18,097.48 $81,902.52
2 $81,902.52 $23,097.48 $4,095.12 $19,002.36 $62,900.16
3 $62,900.16 $22,665.98 $2,516.01 $20,149.97 $42,750.19
4 $42,750.19 $22,665.98 $1,710.01 $20,955.97 $21,794.22
5 $21,794.22 $22,665.98 $871.76 $21,794.22 $0.00
Exercise Questions for Chapter 3 Exercise Question 1
The pavement of a road requires $400,000 per year to maintain. The feasibility of a new pavement is being considered for reducing maintenance costs. If the new pavement needs no maintenance in the first three years, then $200,000 per year for the next seven years, and then $400,000 per year thereafter, what is the immediate expenditure for the new pavement that is justifiable? (Assume a discount rate of 10% p.a.).
Exercise Question 2
A contractor borrowed $500,000 from a bank to buy earth-moving equipment with an estimated service life of 10 years. The bank charged the contractor 12% interest p.a. and required him to pay back the loan in 10 years’ time.
a) Assuming that the contractor paid back the bank in 10 equal instalments (once every year), calculate the amount of each end-of-year payment.
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Compound Interest, Net Present Value (NPV), Equivalent Annual Cost and Loan Redemption b) The contractor at the end of year 4 wished to make an early redemption (i.e. pay all the
money that he owed the bank). How much should he pay?
c) The bank negotiated with the contractor and reduced the interest rate to 10% p.a. at the beginning of the 5th year in order to attract the contractor to stay borrowing. What would be the contractor’s repayment schedule if he chose to pay back the bank in the form of six uniform payments from the end of years 5 to the end of year 10?
d) If the bank changed the interest rate back to 12% p.a. at the beginning of the 8th year, what would be the amount of the contractor’s last payment (i.e. payment at the end of year 10) if he kept on paying the bank the same instalment as calculated in (c) above at the end or years 8 and 9?
Exercise Question 3
There are two alternatives to construct a storage house. Both serve the purpose of allowing construction materials to be stored in the house. However, due to different construction methods (one is made of wood and the other made of bricks), different life spans and cash flow patterns are associated with each alternative as follows:
Alternative 1 (wood) Alternative 2 (bricks)
Life 10 years 15 years
Initial capital cost $900,000 $1,300,000
Operation and maintenance cost $80,000 p.a. $20,000 p.a.
Assuming the discount rate to be 16% p.a., choose the better alternative by:
a) the present value method, and b) the equivalent annual cost method.
(Hints: compare the alternatives based on the same number of years, i.e. 30 years)
Exercise Question 4
We have seen from Section 3.5 (or Example 3.6) of Chapter 3 that no matter what methods we use to pay back a loan, the sum of the present values of the payments is a constant value, equal to the principal amount borrowed. Prove mathematically to show that such a phenomenon is always true.
Exercise Question 5
We have seen from Section 3.6 (or Example 3.7) of Chapter 3 that the number is zero at the bottom- right corner of the table in Part a) of the example. Prove mathematically to show that this is always true for all cases in general.
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Internal Rate of Return (IRR) and the Differences between IRR and NPV
